• 2022-06-06
    这里介绍一些与随机变量x和xz的协方差有关的关系式。[br][/br]我们计算了[tex=13.786x1.357]wiXrGoO8wEpHbGy1Lsyf6bjrlBwLw09D43lTj4xmXPeS7E2H6AdTTc31MUmoTD0U[/tex]的方差。如果[tex=7.857x1.357]YRHgHmN/yZW92ECOHesamrZfkPHdXXZbcDre+I8M3j1TECwf1vI0zF7Yz6TSJVhYr+IHre92vyKXBaAAri6TFg==[/tex],上面的结论一—当[tex=3.143x1.0]ZxR9r3QjH1bAu/oUyVc20A==[/tex]时,X的方差最小,是否还成立?
  • 此时,有:[tex=23.643x4.643]Uhu2/1bhRl3dw2iQmc0iyUSWAm9y+aO3PY1ofjNwi8d0apPMVnEKoKYPRL+XLUwkFF8fl/i8UCsGCFKAG68ncM2HCltAH4gqzMxoANLpGFOXh0NbrqLPqAXWrmF5UDLhIRbJFaWaEzjRvEz0L/kMyI670bcWheed9ujS6cKTQg/QDCNWCvH0BgrF+znTz6ZzNdMBcutd6EFC5X3SNSGu+Y4b6wFDJOnm4T/1Rk7olvgIDfcODTjHI6LQqp1uCjG3IjsUwgxLFuvRAlhaPgVzxw0NdERBMYAhnZNVEkfhXYGfng48Dcumxq9auVb42RknK2S00JhSiLaH7xdo9rPJSoL6OqZYG2I/mJQ7uer8rBDhLjzzGaik7wuG0TKzpESTJAG04kNwsKzZsBPLNKRQNZ+H+KXC27iA9UexyYH62TMJ+IiyyhKatYm75pFsKSd32R4pNIcUbBsm9oyCnzksXq/CNNQuZijEuMbUxA1aCIU=[/tex]一阶条件为:[tex=25.571x1.5]62RacqMaNGGhYC1c+s6PRvbFe2W9J35RX4hXte6J1NYZmW+/AzxdryJSxbiwSpgjndUL6G6akUV2iaoG4rFS0SbpNt34ATlieP+2+JTi2wLuGfNIfGB/sQZLQ07PdUTQlygWetyv1lAvfKs9ms3B0+tJqyrB4ykiHaeyA64czuqY8F1NgQnTPs5qGU6Pw84n[/tex]解得:[tex=19.643x2.714]leU9NT8m0d79ciOlbgSIHO6ad8AzwDCkyM+RBsSGWmGe2g7/7gab7EN69OEGprsCY9X+IWqolHdetN6UAfBiZwNuoyg/38M3ZRt6/N4T10MgUGMGnreqx3HN7kX70wuNI7Z1Oxy60EQxerYaPqD8fR7nc61ZBWYyCVZc4x74MwFlR2Vl1AkIJ+jhcqFqPxqV2jqes88NuwZZOI4odharRwtrEebwUEQs/Bb5MsOEAXPCdGrOiNgnz0ZDIGZcHuE3oqqYnHIxBfdNzhmfC/Gac5rDbzoigpBvkFWOX8/Qhj0=[/tex],因此结论依然成立。

    举一反三

    内容

    • 0

      >>>x= [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9]>>>print(x.sort()) 语句运行结果正确的是( )。 A: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] B: [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9] C: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] D: ['2', '4', '0', '6', '10', '7', '8', '3', '9', '1', '5']

    • 1

      设随机变量X的分布律为P{X=-1}=1/6, P{X=0}=1/3, P{X=1/2}=1/6, P{X=1}=1/12, P{X=2}=1/4, 则E(X²)= ( ). A: 1/3 B: 2/3 C: 31/24 D: 4/3

    • 2

      以下创建数组的方式错误的是() A: shortx[];x={1,2,3,4,5,6}; B: shortx[]=newshort[6];x[0]=9;x[1]=8;x[2]=7;x[3]=6;x[4]=5;x[5]=4; C: shortx[]=newshort[6];intlen=x.length;for(inti=0;i

    • 3

      以下程序的输出结果是() main( ) { int i , x[3][3]={9 , 8 , 7 , 6 , 5 , 4 , 3 , 2 , 1} , *p=&x[1][1] ; for(i=0 ; i<4 ; i+=2) printf("%d " , p[i]) ;

    • 4

      ‌下面说法错误的是( )。‌‌知识点:列表推导式‌ A: dict([(x, x**2) for x in range(6)]) 创建的字典是{0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25} B: [[x*3+y for y in range(1,4)] for x in range(3)] 创建的是二维列表 [[1, 2, 3], [4, 5, 6], [7, 8, 9]] C: number = [-2, 4, 6, -5]string = 'ab'z = [(i, j) if i>0 else (-i, j) for i in number for j in string]这段代码创建的列表为[(2, 'a'), (2, 'b'), (4, 'a'), (4, 'b'), (6, 'a'), (6, 'b'), (5, 'a'), (5, 'b')] D: ' '.join([i for i in range(1,11)])的运算结果为字符串'1 2 3 4 5 6 7 8 9 10'